Question

having issues....thought i understood it, but i keep getting the wrong answer. can someone explain what I'm doing wrong here?

parametric curve

x=5t+lnt
y=6t-lnt

find d^2y/dx^2

Answer 1

so far i have dy/dx = 6t-1/5t+1

Answer 2

d^2y/dx^2=\[\frac{ d/dt(\frac{ 6t-1 }{ 5t+1 } }{ 5t+1 }\]

Answer 3

d/dt = 30t+6-30t+5/(5t+1)^2=11/(5t+1)^2

all over another 5t+1 giving 11/(5t+1)^3

Answer 4

but webwork says it's wrong...what did i do?

Answer 5

\(\large \frac{d^2y}{dx^2} = \frac{d/dt (dy/dx)}{dx/dt}\)

Answer 6

denominator = 5 + 1/t and not just 5t+1
right ?

Answer 7

\(\large \frac{d^2y}{dx^2} = \frac{d/dt (\frac{6t-1}{5t+1})}{5 + 1/t}\)

Answer 8

because dx/dt is 5+1/t

Answer 9

http://www.youtube.com/watch?v=7ANhq51wjWM

Answer 10

so that video explaining it is wrong?

Answer 11

first slide in that video looks correct oly ?
it uses the same formula we're using eh ?

Answer 12

it doesn't have the extra t...i did the steps the same way and there is no extra t...it just divides everything by dx which is 5t+1. you guys are saying divide by 5t+1/t

Answer 13

what did you get as dx/dt =... ?

Answer 14

\(\large \frac{dx}{dt} = 5 + \frac{1}{t}\)

Answer 15

once u have dy/dx,
to take the second derivative, you need to do it in parametric way..

find dy'/dt,
and divide it wid dx/dt

** formula above

Answer 16

the derivative of x is 5+1/t or 5t+1. all these dy, dx, dt, confuses the tar out of me. d/dx is derivative of x, right?

Answer 17

**the derivative of x is 5+1/t or 5t+1

who ate the bottom t ? :)

Answer 18

5 + 1/t = (5t+1)/t

Answer 19

5+1/t = (5t+1)/t

Answer 20

no one...it's just 5+1/t...times the whole thing by t to make it simpler to deal with gets you 5t+1

Answer 21

5+(1/t) *t is 5t+1

Answer 22

ugh by that logic, 1+1/3 is same as 3*1 + 1 is it

Answer 23

why ...*t ?

Answer 24

to get rid of the fraction...(5+1/t) *t distribute the t = 5t+t/t or 5t+1

Answer 25

to simplify and get rid of fraction, u need to multiply wid 1 = t/t
both top and bottom u need to multiply wid t ok

Answer 26

\(\large 5 + \frac{1}{t}\)


\(\large 5\frac{t}{t} + \frac{1}{t}\)


\(\large \frac{5t+1}{t}\)

Answer 27

sorry, here you cant get rid of fraction as such...

Answer 28

looks u have applied the formula correctly,
just made an error during simplifying fracitons :o

Answer 29

i got that rewrite directly from a derivative calculator....

Answer 30

it said 6-1/t / 5+1/t was the same as 6t-1/5t+1.

Answer 31

its same,
but in ur derivative calculator page, you're just finding \(\large \frac{d}{dt} (\frac{dy}{dx})\)

Answer 32

you need to divide watever u get above by dx/dt ok

Answer 33

i'm so confused....so it's the derivative of the dy/dx times what over what? i don't understand what d/dt(dy/dx)/dx/dt...that's just too many d's t's x's and y's...

Answer 34

lets start from scratch.
it wont take more than 10 minutes to digest this for u

Answer 35

yeah, i understand "derivative" and f prime and all that, but these notations confuse me. thank you for being patient with me, @rational

Answer 36

\(x=5t+\ln t\)

\(y=6t-\ln t \)

\(\large \frac{dx}{dt} = 5 + \frac{1}{t}\)

\(\large \frac{dy}{dt} = 6 - \frac{1}{t}\)



\(\huge \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

\(\huge ~~~= \frac{6 - \frac{1}{t}}{5 + \frac{1}{t}}\)

\(\huge ~~~= \frac{6t-1}{5t+1}\)

Answer 37

this far, you have worked already.

Answer 38

it takes time to get use to, dont wry... two more problems like these... and u wil feel confy :)

Answer 39

next lets find \(\large \frac{d^2y}{dx^2}\)

Answer 40

\(\huge \frac{d^2y}{dx^2} = \frac{\frac{d}{dt} (\frac{dy}{dx})}{\frac{dx}{dt}}\)

Answer 41

ok, with you so far.

Answer 42

plug things in above formula

Answer 43

\(\huge \frac{d^2y}{dx^2} = \frac{\frac{d}{dt} (\frac{dy}{dx})}{\frac{dx}{dt}}\)


\(\huge \frac{d^2y}{dx^2} = \frac{\frac{d}{dt} (\frac{6t-1}{5t+1})}{5+\frac{1}{t}}\)

Answer 44

find the derivative on numerator and simplify.

Answer 45

u knw (f/g)' formula right

Answer 46

ok, so (11/(5t+1)^2) / (5+1/t)

Answer 47

thats correct !
you may simplify more if u want

Answer 48

\(\huge \frac{d^2y}{dx^2} = \frac{\frac{d}{dt} (\frac{dy}{dx})}{\frac{dx}{dt}}\)


\(\huge \frac{d^2y}{dx^2} = \frac{\frac{d}{dt} (\frac{6t-1}{5t+1})}{5+\frac{1}{t}}\)

\(\huge \frac{d^2y}{dx^2} = \frac{ \frac{11}{(5t+1)^2}}{5+\frac{1}{t}}\)


\(\huge \frac{d^2y}{dx^2} = \frac{ \frac{11}{(5t+1)^2}}{\frac{5t+1}{t}}\)

\(\huge \frac{d^2y}{dx^2} = \frac{11t}{(5t+1)^3}\)

Answer 49

thanks. i got the right answer. finally. test friday.

Answer 50

:) good luck !